Video: Garfield’s Proof of the Pythagorean Theorem

We look at James Garfield’s, the 20th American president, geometric proof of the Pythagorean theorem. This proof involves taking a copy of the triangle and placing it to create a trapezoid and then comparing two ways of calculating the area of that trapezoid.


Video Transcript

In this video, we are going to look at a particular proof of the Pythagorean theorem by a man named James Garfield. Now what’s interesting about James Garfield is he wasn’t actually primarily a mathematician; he was a politician and he was in fact the twentieth president of the United States of America. But in amongst all of that, he still managed to come up with this proof of the Pythagorean theorem and he did that in 1876. And that’s what we’re going to look at in this video.

Before we look at the proof, let’s just remind ourselves what the Pythagorean theorem tells us. And remember it’s all to do with right-angled triangles and specifically to do with the relationships between the lengths of the three sides. So the statement of the theorem here, “the square of the hypotenuse is equal to the sum of the squares of the two shorter sides.” The hypotenuse remember is the longest side of the right-angled triangle. And so if we square that, we get the same result as if we square the other two sides and add them together. The diagram on the left-hand side of the screen just shows this pictorially. So we’ve got each of the sides of that triangle with the squares drawn on them. And Pythagorean theorem tells us that 𝑎 squared plus 𝑏 squared is equal to 𝑐 squared, where 𝑎 and 𝑏 are the two shorter sides and 𝑐 is the hypotenuse.

So let’s have a look at Garfield’s proof. So I’m starting off with the right-angled triangle and I’ve labelled the three sides as 𝑎, 𝑏, and 𝑐. What Garfield did was he took an identical copy of this right-angled triangle and he put it just above it, so a rotated copy of the exact same triangle in this location here. And then he joined up this point here and this point here to form a trapezium or a trapezoid. So the trapezoid we’re looking at is this blue line and then the base of the shape here, the top here, and then the vertical side down here.

Now what Garfield did was he looked at two alternative ways of calculating the area of this trapezoid. And the first of those is using this sort of general formula that we use in order to calculate the area of a trapezium or a trapezoid, which is to add together the two parallel sides, so in this case that’s going to be 𝑎 and 𝑏, and then divide that by two. So 𝑎 plus 𝑏 over two which gives an average of those two sides. We then multiply by the distance between those two sides. So the distance between the two sides is the vertical height of this trapezium, which is 𝑎 plus 𝑏. And that’s just the general method for calculating the area of a trapezium.

You can then go through the algebra of simplifying this, so expanding brackets. And if you do that, you will get 𝑎 squared plus 𝑏 squared plus 𝑎𝑏 plus 𝑎𝑏 all over two, which simplifies to 𝑎 squared plus 𝑏 squared over two plus 𝑎𝑏. Because I had two of them and if I divide by two, then it cancels out to just one. So this is one expression for the area of that trapezium or trapezoid.

We can then think about an alternative way to calculate this area by looking at the area of the three individual triangles. So we have triangle one, triangle two, and triangle three. Now triangle one and triangle two, they’re both right-angled triangles with sides of 𝑎 and 𝑏. So the area of each of those is going to be 𝑎 multiplied by 𝑏 divided by two. And I’ve got two of them, so I’m gonna have 𝑎𝑏 over two twice.

Now I have to think about the area of triangle three and it looks like a right-angled triangle, and it in fact is. But we just need to convince ourselves why that’s the case. If I give this angle here a letter 𝜃, then this angle here is also 𝜃 because the two triangles were identical copies of each other. And then using the fact that angles in a triangle have to add up to a hundred and eighty degrees, I can deduce that this angle here must be ninety minus 𝜃 because I’ve already got one right angle which is ninety degrees and I’ve got 𝜃. So what’s left over is ninety minus 𝜃.

Finally, to work out the size of the angle in triangle three, I need to use the fact that angles on a straight line add up to a hundred and eighty degrees. So I’ve already got an angle of 𝜃 and I’ve got an angle of ninety minus 𝜃, so together those two angles add up to ninety degrees, which means the remaining part — this part here — must also be ninety degrees in order to make that hundred and eighty degrees on a straight line. So now I’ve proven that triangle three is in fact a right-angled triangle. So to work out its area, I can do base times height over two. So that’s gonna be 𝑐 multiplied by 𝑐 over two or just 𝑐 squared over two. Now if I simplify this result, I have 𝑎𝑏 over two plus 𝑎𝑏 over two. So that’s just going to become 𝑎𝑏. So I’m left with 𝑐 squared over two plus 𝑎𝑏. And there is our second version of the area of this trapezium.

The final step in this proof is then to put these two areas equal to each other. Because if they’re describing the same shape, then they must both give the same result. So I can take my two different expressions for the area and put them equal to each other. Now what you notice is they both have plus 𝑎𝑏 — this term here. So that will cancel out directly and I’m left with 𝑎 squared plus 𝑏 squared over two equals 𝑐 squared over two. So I can multiply both sides of this equation by two, which will have the effect of eliminating those twos from the denominator. And what I will be left with is the statement 𝑎 squared plus 𝑏 squared is equal to 𝑐 squared, which is of course the statement of the Pythagorean theorem. So there you have Garfield’s proof of the Pythagorean theorem, a really nice geometric proof that enables us to see for ourselves that this theorem is true.

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